You have found the following ages (in years) of 4 seals. Those seals were randomly selected from the 48 seals at your local zoo: $ 9,\enspace 18,\enspace 9,\enspace 9$ Based on your sample, what is the average age of the seals? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 48 seals, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $4$ samples and divide by $4$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\overline{x}} = \dfrac{9 + 18 + 9 + 9}{{4}} = {11.3\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {5.29} + {44.89} + {5.29} + {5.29}} {{4 - 1}} $ {s^2} = \dfrac{{60.76}}{{3}} = {20.25\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{20.25\text{ years}^2}} = {4.5\text{ years}} $ We can estimate that the average seal at the zoo is 11.3 years old. There is also a standard deviation of 4.5 years.